Theory and applications of finite groups, by G.A. Miller, H. F. Blichfeldt [and] L. E. Dickson.

352

LINES ON A CUBIC SURFACE

[CH. XIX

But there is an additional bitangent to C4, making 28 in all.
In fact, the plane L=0 passes through (y), in view of Euler's
theorem (4), ~ 173. Hence L=0 intersects T4, given by (3),
in two pairs of coincident lines, the intersections of L = 0, Q = 0.
Thus L=0 is a bitangent plane to T4.
Before we can conclude that the general plane quartic
curve has exactly 28 bitangents, we must show that, conversely,
any given quartic curve C4 is the intersection of the plane
U of the curve with the tangent cone to a suitably chosen cubic
surface f at a point P on it.
Let x, y, z be the homogeneous coordinates of a point in the
plane U referred to a triangle of reference whose side z=0 is
z=O
FIG. 1.
FIG. 17.
one of the bitangents to C4 and whose sides x=0 and y =0 are
any lines through the points of contact of this bitangent. Then
the equation of C4 reduces to x2y2 =0 when z =0 and hence is
(4) zo-x2y2 =0,
where f is a cubic form in x, y, z.
Let P be any point not in the plane U of C4. As the tetrahedron of reference for homogeneous coordinates x, y, z, u of
points in space, take that determined by the plane U and the
planes through P and the sides of our triangle of reference.
Then the desired cubic surface is
(5) f= ~+4uxy+4u2z =O.
In fact, we shall proceed with this f as we did with the
general f and find its tangent cone with the vertex P= (O, 0,
0, 1) in place of (y). Now
af = a+4uy, f = _ +4ux,
ax ax ay ay
af _= a+4u2 af 4xy + 8uz.
az az au